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valentinak56 [21]
3 years ago
10

Perpendicular slope to 3x-y=4

Physics
1 answer:
Mashcka [7]3 years ago
4 0
Perpendicular slope would be 1/3. so the equation will be Y=1/3x -4

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Did I do these questions correctly?
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Yes, they seem right to me.
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Which of the following is NOT true about energy conversion devices?
konstantin123 [22]

Answer:

D

Explanation:

Option A, B and C are true about energy conversion devices, only option D is NOT true about energy conversion devices.

5 0
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Explain how evaporation cools a liquid​
MAVERICK [17]

When evaporation occurs liquid absorbs heat from the surroundings to get converted to its vapour form as a result, there is an overall decrease in the heat leading to cooling of the liquid.

Hope that this was helpful :)

8 0
2 years ago
An object with more mass has more kinetic energy than an object with less
Ber [7]

Explanation:

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K=\dfrac{1}{2}mv^2

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4 0
3 years ago
At what rate would heat be lost through the window if you covered it with a 0.750 mm-thick layer of paper (thermal conductivity
igor_vitrenko [27]

The given question is incomplete. The complete question is as follows.

A picture window has dimensions of 1.40 m, 2.50 m and is made of glass 6.00 mm thick. On a winter day, the outside temperature is -15^{o}C, while the inside temperature is a comfortable 21.0^{o}C.

Part A

At what rate is heat being lost through the window by conduction?

Express your answer using three significant figures.

Part B

At what rate would heat be lost through the window if you covered it with a 0.750 mm-thick layer of paper (thermal conductivity 0.0500 W/m?K)?

Express your answer using three significant figures.

Explanation:

Formula for rate of heat transferred through single thick plane of glass is as follows.

       (\frac{Q}{\Delta t)}_{single} = \frac{A(T_{h} - T_{c})}{\frac{L_{ghss}}{K_{ghss}}}

                               = \frac{1.4 \times 2.5 m^{2} \times 36^{o}C}{\frac{6.0 mm}{0.8 W/m K}}

                           = 16.8 \times 10^{3} W

When window is covered by paper then the rate of heat transfer is as follows.

(\frac{Q}{\Delta t)}_{single} = \frac{A(T_{h} - T_{c})}{\frac{L_{ghss}}{K_{ghss}} + \frac{L_{paper}}{K_{paper}}}

                 = \frac{(1.4 \times 2.5 m^{2} \times 36^{o}C}{\frac{6.0 mm}{0.8 W/m K} + \frac{0.75 mm}{0.05 W/m K}}

                 = 5.6 \times 10^{3} W

Thus, we can conclude that heat lost is 5.6 \times 10^{3} W.

3 0
3 years ago
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